I don't understand this step in the proof for the probability of coprimality

Question

On wikipedia for the probability that two random numbers are coprime they give the following line $\left(\prod\frac{1}{1-p^{-2}}\right)^{-1}=\frac{1}{\zeta(2)}$ where $1-\frac{1}{p^2}$ is the probability that at least one of the two numbers are not divisible by $p$ which is the arbitrary prime number. Could someone give an explanation on how this equation is derived, I can't find it from googling the full proof of coprimality.

Answer 1

We want to prove that $\left(\prod\frac{1}{1-p^{-2}}\right)^{-1}=\frac{1}{\zeta(2)}$

Recall that \begin{align*} \zeta(2) &= \sum_{n=1}^{\infty} \frac{1}{n^2} \\ &= 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \end{align*}

To arrive at our intended product, we must first sieve out all $\{ z\in\mathbb{Z_+}:z \neq 1 \} $ by removing multiples of primes along with the primes themselves (by fundamental theorem of arithmetic). We can do this by multiplying $\zeta(2)$ by $1 - \frac{1}{p^2}$ for all primes $p$. For example, for $p=2$ \begin{align*} \zeta(2) &= 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \\ \zeta(2) (1 - \frac{1}{2^2}) &= (1 - \frac{1}{2^2})( \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2}) \\ &= \zeta(2) + (-\zeta(2) \times \frac{1}{2^2})) \end{align*}

Evaluating $(-\zeta(2) \times \frac{1}{2^2}))$ \begin{align*} (-\zeta(2) \times \frac{1}{2^2})) &= (-\frac{1}{2^2} \times \zeta(2)) \\ &= -\frac{1}{2^2}( 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2}) \\ &= -\frac{1}{2^2} - \frac{1}{4^2}- \frac{1}{6^2}\cdots \end{align*}

As such, by combining the two series: \begin{align*} \zeta(2) + (-\zeta(2) \times \frac{1}{2^2})) = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \cdots \end{align*}

Similarly for p=3, \begin{align*} \zeta(2)[(1 - \frac{1}{2^2})(1 - \frac{1}{3^2}) ] &=\\ &= \zeta(2) + (-\zeta(2) \times \frac{1}{2^2})) + (-\zeta(2) \times \frac{1}{3^2})) \\ &= 1+ \frac{1}{5^2} + \frac{1}{7^2} + \cdots \end{align*}

When all $p\leq n$ have been sieved, we have the following: \begin{align*} \zeta(2)[ (1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{4^2})\cdots(1 - \frac{1}{p^2})] &= 1 \\ [ (1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{4^2}) \times \cdots \times (1 - \frac{1}{p^2})] &= \frac{1}{\zeta(2)} \\ \prod(1 - \frac{1}{p^2})] &= \\ \implies \left(\prod\frac{1}{1-p^{-2}}\right)^{-1}&= \frac{1}{\zeta(2)} \blacksquare \end{align*}

Note that this is just a special case of an Euler product.